Sketch the curve y=4-(x+3)^2, showing the points where the curve crosses the x-axis and any minimum or maximum points.

This equation rearranges to give -y=(x+3)^2-4, which is very similar to our curve y=(x+3)^2-4 from before. In fact, replacing y with -y in an equation is equivalent to reflecting the curve through the x-axis. We then take the points (-5,0), (-1,0) and (-3,-4) from before and replace y with -y, giving (-5,0), (-1,0) and (-3,4). We have found where the new curve crosses the x-axis and its minimum/maximum. The graph is an inverted u-shape since we have a -x^2 in the equation so (-3,4) is a maximum point.

JI
Answered by Jonny I. Maths tutor

3563 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the equation: x^2 +8x + 12 = 0


How do you solve this problem?


How do you work out compound interest?


What is the equation of the straight line perpendicular to the midpoint of the straight line that passes through (0,5) and (-4,7)?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences